1. IP address = 192.168.1.102, port number = 1161
a.
2. IP address = 128.119.245.12, port number = 80
a.
3. N/A
4. Sequence number = 0 – I can identify this segment as a SYN because it identifies it as a
SYN in the “Info” column and also because the SYN flag is set in this segment
a.
5. Sequence number = 0, Acknowledgement number = 1 – This value was determined by
adding 1 to the initial sequence number of the SYN segment sent to gaia.cs.umass.edu –
Additionally, the ACK value for a SYNACK segment is equal to the sequence number of
the next ACK segment – I can identify this segment as a SYNACK segment b/c Wireshark
states that its SYN and ACK flags are set
a.
6. Sequence number = 1
a.
7. First six segments of TCP connection:
a.
i. Segment 1: Sequence number = 1, time = 0.026477
1. Ack was received at 0.053937
2. RTT = ACK received time – packet sent time = 0.02746
3. EstimatedRTT = (1 - .125) * EstimatedRTT + .125 * SampleRTT
a. Since this is the first segment in the connection, let
EstimatedRTT = RTT = 0.02746
ii. Segment 2: sequence number = 566, time = 0.041737
1. Ack was received at 0.077294
2. RTT = 0.035557
3. EstimatedRTT = (1 - .125) * 0.02746 + .125 * 0.035557 = 0.0285
iii. Segment 3: sequence number = 2026, time = 0.054026
1. Ack was received at 0.124085
2. RTT = 0.070059
3. EstimatedRTT = 0.0337
iv. Segment 4: sequence number = 3486, time = 0.054690
1. Ack was received at 0.169118
2. RTT = 0.11443
3. EstimatedRTT = 0.438
v. Segment 5: sequence number = 4946, time = 0.077405
1. Ack was received at 0.217299
2. RTT = 0.13989
3. EstimatedRTT = 0.0558
vi. Segment 6: sequence number = 6406, time = 0.078157
1. Ack was received at 0.267802
2. RTT = 0.18964
3. EstimatedRTT = 0.0725
8.
a.
b.
c.
d.
e.
f.
Segment 1: 565 bytes
Segment 2: 1460 bytes
Segment 3: 1460 bytes
Segment 4: 1460 bytes
Segment 5: 1460 bytes
Segment 6: 1460 bytes
9. The minimum amount of available buffer space that I observed was 5480 bytes – from
here, the window size grows steadily to a maximum buffer size of 62780 bytes – I never
observed the sender being throttled due to the lack of receiver buffer space
a.
10. There are no retransmitted segments in the trace file. I know this because I checked the
sequence numbers of all the TCP segments in the trace file and they were all in order,
increasing.
11. The receiver typically acknowledges about 1460 bytes of data (reference question 8). **
a. How much data does the receiver typically acknowledge in an ACK? Can you
identify cases where the receiver is ACKing every other received segment (see
Table 3.2 on page 250 in the text).
12. Throughput = (# bytes transferred during TCP connection) / total transmission time =
30.222 KB / sec
a. # bytes transferred during the TCP connection = acknowledged sequence number
of last ACK – sequence number of first TCP segment = 164091 – 1 = 164090 bytes
b. Total transmission time = time of last ACK – time of first TCP segment = 5.455830
– 0.026477 = 5.4294
13. The slow start phase lasts from 0-.15 seconds. Around here, congestion avoidance takes
over. One way the measured data differs from idealized behavior of TCP we learned
about is that it only uses a small fraction of the buffer window size.
a.
14. N/A, wasn’t able to capture that information.